/*
Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
*/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode dummy(-1); dummy.next = head;
        ListNode *p1 = &dummy, *p2 = &dummy;
        while (n--) p2 = p2->next;
        for (; p2 && p2->next; p1=p1->next, p2=p2->next);
        ListNode *to_delete = p1->next;
        // p1->next points to node to be deleted
        p1->next = p1->next->next;
        delete to_delete;
        return dummy.next;
        
    }
};

#if 0
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *p1 = head, *p2 = head;
        while (n--) p2 = p2->next;
        for (; p2 && p2->next; p1=p1->next, p2=p2->next);
        ListNode *to_delete = (p2)?p1->next:head;
        if (p2) p1->next = p1->next->next;
        else head = head->next; // remove first element
        delete to_delete;
        return head;
    }
};
#endif
